Coupled Oscillations
Simple Coupled Oscillations
Let’s say we have two objects, $m_{1}$ and $m_{2}$, connected by two springs as shown in the above figure. Let the distance from the equilibrium point to object $m_{1}$ be $x_{1}$, and to object $m_{2}$ be $x_{2}$. The restoring force exerted by a spring on an object is the product of the spring constant and the stretch (or compression) of the spring, so the force exerted by spring 1 on object 1 is $-k_{1}x_{1}$. Since spring 2 is compressed, it pushes object 1 to the left, so the force exerted by spring 2 on object 1 is $-k_{2}(x_{1}-x_{2})$. Therefore, the equation of motion for object 1 is as follows:
$$ \begin{align*} && m_{1}\ddot{x_{1}} &=-k_{1}x_{1}-k_{2}(x_{1}-x_{2}) \\ \implies && \ddot{x_{1}}+ \frac{k_{1}+k_{2}}{m_{1}}x_{1}-\frac{k_{2}}{m_{1}}x_{2}&=0 \end{align*} $$
Object 2 is pulled to the left by the extension of spring 2, so the force exerted by spring 2 on object 2 is $-k_{2}(x_{2}-x_{1})$. Therefore, the equation of motion for object 2 is as follows:
$$ \begin{align*} && m_{2}\ddot{x_{2}} &= -k_{2}(x_{2}-x_{1}) \\ \implies && \ddot{x_{2}}+\frac{k_{2}}{m_{2}}x_{2}-\frac{k_{2}}{m_{2}}x_{1} &=0 \end{align*} $$
Hence, the equations of motion for the system illustrated above can be represented by the following system of coupled differential equations:
$$ \left\{ \begin{align*} \ddot{x_{1}}+ \frac{k_{1}+k_{2}}{m_{1}}x_{1}-\frac{k_{2}}{m_{1}}x_{2}&=0 \\ \ddot{x_{2}}+\frac{k_{2}}{m_{2}}x_{2}-\frac{k_{2}}{m_{2}}x_{1} &=0 \end{align*} \right. $$
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Coupled Oscillations with Three Springs
Object $m_{1}$ is acted upon by both spring 1 and spring 2. Calculating as before, these forces are respectively $-k_{1}x_{1}$ and $-k_{2}(x_{1}-x_{2})$. Therefore, the equation of motion for object 1 remains the same.
$$ \ddot{x_{1}}+ \frac{k_{1}+k_{2}}{m_{1}}x_{1}-\frac{k_{2}}{m_{1}}x_{2}=0 $$
Object $m_{2}$ is acted upon by both spring 2 and spring 3. These forces are respectively $-k_{2}(x_{2}-x_{1})$ and $-k_{3}x_{2}$. Therefore, the equation of motion for object 2 is as follows:
$$ \begin{align*} && m_{2}\ddot{x_{2}} &= -k_{2}(x_{2}-x_{1})-k_{3}x_{2} \\ \implies && \ddot{x_{2}}+\frac{k_{2}+k_{3}}{m_{2}}x_{2}-\frac{k_{2}}{m_{2}}x_{1} &=0 \end{align*} $$
The overall equation of motion for the system can be represented by the following system of coupled differential equations:
$$ \left\{ \begin{align*} \ddot{x_{1}}+ \frac{k_{1}+k_{2}}{m_{1}}x_{1}-\frac{k_{2}}{m_{1}}x_{2}&=0 \\ \ddot{x_{2}}+\frac{k_{2}+k_{3}}{m_{2}}x_{2}-\frac{k_{2}}{m_{2}}x_{1} &=0 \end{align*} \right. $$
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